Question

Find the value of \[x\] for which \[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)\] is satisfied.
(a) \[\dfrac{1}{2}\]
(b) 1
(c) 0
(d) \[-\dfrac{1}{2}\]

Answer 1

Hint: In this question, we are given with the equation \[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)\].
Now we will use the identity of the trigonometric function say \[\sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2}-\theta \right)\]. Also we will use the property that if \[\cos x=\cos y\], then we have that \[x=2n\pi \pm y\] for integers values \[n\]. Using these properties of trigonometric functions, we will get our desired answer.

Complete step by step answer:
We are given with the equation \[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right).............(1)\].
Now since we know that for any \[\theta \in \left[ 0,2\pi \right]\], \[\sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2}-\theta \right)\].
Therefore we can have
\[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( \dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right) \right).............(2)\]
Now on substituting the value of equation (2) in equation (1), we will have
\[\cos \left( \dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right).............(3)\]
Since we also know the property of the trigonometric function that if \[\cos a=\cos b\], then we have that \[a=2n\pi \pm b\] for all integers values \[n\].
Now on comparing equation (3) with \[\cos a=\cos b\], we have that
\[a=\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)\] and
\[b={{\tan }^{-1}}x\]
Therefore using the property that if \[\cos a=\cos b\], then we have that \[a=2n\pi \pm b\] for all integers values \[n\], we will have
\[\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)=2n\pi \pm {{\tan }^{-1}}x\] for all integers values \[n\].
Now since the above equation is true all the integer values of \[n\], hence it is also true for \[n=0\].
Now on substituting the value \[n=0\] in the above equation \[\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)=2n\pi \pm {{\tan }^{-1}}x\], we will have
\[\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)=0\pm {{\tan }^{-1}}x\]
This implies that
\[\dfrac{\pi }{2}={{\cot }^{-1}}\left( 1+x \right)\pm {{\tan }^{-1}}x............(4)\]
Now since we know that \[{{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x\] for all values of \[x\].
Hence we can have \[{{\cot }^{-1}}\left( 1+x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1+x \right)\]
Therefore on substituting the value \[{{\cot }^{-1}}\left( 1+x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1+x \right)\] in the above equation (4), we get
\[\dfrac{\pi }{2}=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1+x \right)\pm {{\tan }^{-1}}x\]
Therefore we have
\[{{\tan }^{-1}}\left( 1+x \right)=\pm {{\tan }^{-1}}x\]
Now since we know that if \[\pm {{\tan }^{-1}}x={{\tan }^{-1}}\left( \pm x \right)\]
Therefore we get
\[{{\tan }^{-1}}\left( 1+x \right)={{\tan }^{-1}}\left( \pm x \right)\]
Also if \[{{\tan }^{-1}}x={{\tan }^{-1}}y\], then \[x=y\].
Therefore form \[{{\tan }^{-1}}\left( 1+x \right)={{\tan }^{-1}}\left( \pm x \right)\], we get that
\[1+x=\pm x\]
Now since we cannot have \[1+x=x\] for any value of \[x\], hence we must have
\[1+x=-x\]
This implies that
\[\begin{align}
  & -2x=1 \\
 & \Rightarrow x=-\dfrac{1}{2}
\end{align}\]
Therefore we have that for \[x=-\dfrac{1}{2}\] we have that the given equation \[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)\] is satisfied.

So, the correct answer is “Option D”.

Note: In this problem, in order to determine the value of \[x\] for which the given equation \[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)\] is satisfied we can substitute the value \[{{\cot }^{-1}}\left( 1+x \right)=a\] and \[{{\tan }^{-1}}x=b\] to get that \[1+x=\cot a\] and \[x=\tan b\]. Then we have found the values of \[a\] and \[b\] using the properties of \[\cot \] and \[\tan \] function. Then on substituting the values in \[\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)\] we have form a quadratic equation in \[x\]. On solving the same we can get the desired answer.